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3 years ago

A 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s what is the net force on the car

Physics

1 answer:

Ksenya-84 [330]3 years ago

8 0

Answer: 20496N

Explanation:

The formula to calculate the net force will be given as:

Net force = Acceleration × Mass

Since 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s, the net force would be:

= 10248 × 2

= 20496N

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When a flat slab of transparent material is placed under water, the critical angle for light traveling from the slab into water

Novosadov [1.4K]

Answer:

(a) 40.6 degree

Explanation:

When refraction takes place from slab to water, the critical angle is 60 degree.

Use Snell's law

refractive index of water with respect to slab

$\mu _{w}^{s}=\frac{Sin60}{Sin90}$

$\frac{\mu _{w}}{\mu _{s}}=0.866$

$\frac{1.33}{\mu _{s}}=0.866$

μs = 1.536

Now for slab air interface, the critical angle is C.

$\mu _{a}^{s}=\frac{SinC}{Sin90}$

$\frac{\mu _{a}}{\mu _{s}}=\frac{SinC}{Sin90}$

1 / 1.536 = Sin C

C = 40.6 degree

3 0

3 years ago

When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the

nataly862011 [7]

Answer:

Given: V=12 V

I=2.5 mA

Let the resistance be R

By Ohm's Law,

V=IR

12=2.5×10−3R

R=4.8×103 Ω

8 0

2 years ago

A quantity that has only one magnitude

Len [333]

Answer:

scalar quantity

Explanation:

Vector quantities have two characteristics, a magnitude and a direction. Scalar quantities have only a magnitude.

3 0

3 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

3 years ago

A wire 2.80 m in length carries a current of 4.20 A in a region where a uniform magnetic field has a magnitude of 0.260 T. Calcu

klio [65]

Answer:.

F = 3.0576sinθ

For any value of θ < 180

Explanation:

Generally, F = BILsinθ

Where,. F = magnetic force magnitude. B = magnetic Field magnitude.

L = length of wire. I = current

Therefore,

B = 0.260 T, L = 2.80 m

I = 4.20 A

: F = 0.260 × 4.20 × 2.80sinθ

∴ F = 3.0576sinθ

4 0

3 years ago